\newproblem{lay:4_1_5}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 4.1.5}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Determine if the set of all polynomials of the form $p(t)=at^2\quad \forall a\in \mathbb{R}$ are a subspace of $\mathbb{P}_2$.
}{
  % Solution
	Let $H=\{p(t)\in\mathbb{P}_2| p(t)=at^2\}$. We need to show that this set meets the three requirements to be a subspace
	\begin{itemize}
		\item $\mathbf{0}\in H$ \\
		      This is true because for $a=0$ we have $p(t)=0t^2=0$.
		\item Given any two polynomials $p_1(t),p_2(t)\in H$, $p_1(t)+p_2(t)\in H$ \\
		      Assume $p_1(t)=a_1t^2$ and $p_2(t)=a_2t^2$, then
					\begin{center}
						$p_1(t)+p_2(t)=a_1t^2+a_2t^2=(a_1+a_2)t^2$
					\end{center}
					So $p_1(t)+p_2(t)\in H$
		\item Given any polynomial $p(t)\in H$ and $c\in \mathbb{R}$, $cp(t)\in H$
					\begin{center}
						$cp(t)=c(at^2)=(ac)t^2$
					\end{center}
					So $cp(t)\in H$
	\end{itemize}
	Since $H$ meets all properties, $H$ is a subspace of $\mathbb{P}^2$.
}
\useproblem{lay:4_1_5}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
